Q: A child on a swing is 1 metre above the ground at the lowest point and 2 metre above the ground at the highest point. The horizontal speed of the child at the lowest point of the swing is (g = 10 m/s^{2} ) :

(a) 4.5 m/sec

(b) 6.4 m/sec

(c) 7.8 m/sec

(d) 10 m/sec

Ans: (a)

Solution: $ \displaystyle v = \sqrt{2 g (h_2 – h_1)}$

$ \displaystyle v = \sqrt{2 \times 10 (2 – 1)}$

$ \displaystyle v = \sqrt{20} = 4.5 m/s$